Merge pull request #218 from gostool/leetcode1705

Leetcode1705

leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples.go

@@ -0,0 +1,67 @@
+package leetcode
+
+import "container/heap"
+
+func eatenApples(apples []int, days []int) int {
+	h := hp{}
+	i := 0
+	var ans int
+	for ; i < len(apples); i++ {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if apples[i] > 0 {
+			heap.Push(&h, data{apples[i], i + days[i]})
+		}
+		if len(h) > 0 {
+			minData := heap.Pop(&h).(data)
+			ans++
+			if minData.left > 1 {
+				heap.Push(&h, data{minData.left - 1, minData.end})
+			}
+		}
+	}
+	for len(h) > 0 {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if len(h) == 0 {
+			break
+		}
+		minData := heap.Pop(&h).(data)
+		nums := min(minData.left, minData.end-i)
+		ans += nums
+		i += nums
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+type data struct {
+	left int
+	end  int
+}
+
+type hp []data
+
+func (h hp) Len() int           { return len(h) }
+func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
+func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *hp) Push(x interface{}) {
+	*h = append(*h, x.(data))
+}
+
+func (h *hp) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}

leetcode/1705.Maximum-Number-of-Eaten-Apples/1705.Maximum Number of Eaten Apples_test.go

@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1705 struct {
+	para1705
+	ans1705
+}
+
+// para 是参数
+type para1705 struct {
+	apples []int
+	days   []int
+}
+
+// ans 是答案
+type ans1705 struct {
+	ans int
+}
+
+func Test_Problem1705(t *testing.T) {
+
+	qs := []question1705{
+
+		{
+			para1705{[]int{1, 2, 3, 5, 2}, []int{3, 2, 1, 4, 2}},
+			ans1705{7},
+		},
+
+		{
+			para1705{[]int{3, 0, 0, 0, 0, 2}, []int{3, 0, 0, 0, 0, 2}},
+			ans1705{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1705------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1705, q.para1705
+		fmt.Printf("【input】:%v    【output】:%v\n", p, eatenApples(p.apples, p.days))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1705.Maximum-Number-of-Eaten-Apples/README.md

@@ -0,0 +1,124 @@
+# [1705. Maximum Number of Eaten Apples](https://leetcode-cn.com/problems/maximum-number-of-eaten-apples/)
+
+## 题目
+
+There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.
+
+You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.
+
+Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.
+
+**Example 1**:
+
+    Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
+    Output: 7
+    Explanation: You can eat 7 apples:
+    - On the first day, you eat an apple that grew on the first day.
+    - On the second day, you eat an apple that grew on the second day.
+    - On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
+    - On the fourth to the seventh days, you eat apples that grew on the fourth day.
+
+**Example 2**:
+
+    Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
+    Output: 5
+    Explanation: You can eat 5 apples:
+    - On the first to the third day you eat apples that grew on the first day.
+    - Do nothing on the fouth and fifth days.
+    - On the sixth and seventh days you eat apples that grew on the sixth day.
+
+**Constraints:**
+
+- apples.length == n
+- days.length == n
+- 1 <= n <= 2 * 10000
+- 0 <= apples[i], days[i] <= 2 * 10000
+- days[i] = 0 if and only if apples[i] = 0.
+
+## 题目大意
+
+有一棵特殊的苹果树，一连 n 天，每天都可以长出若干个苹果。在第 i 天，树上会长出 apples[i] 个苹果，这些苹果将会在 days[i] 天后（也就是说，第 i + days[i] 天时）腐烂，变得无法食用。也可能有那么几天，树上不会长出新的苹果，此时用 apples[i] == 0 且 days[i] == 0 表示。
+
+你打算每天 最多 吃一个苹果来保证营养均衡。注意，你可以在这 n 天之后继续吃苹果。
+
+给你两个长度为 n 的整数数组 days 和 apples ，返回你可以吃掉的苹果的最大数目。
+
+## 解题思路
+
+贪心算法和最小堆
+
+  - data中的end表示腐烂的日期，left表示拥有的苹果数量
+  - 贪心:每天吃掉end最小但没有腐烂的苹果
+  - 最小堆:构造类型为数组(数组中元素的类型为data)的最小堆
+
+## 代码
+
+```go
+package leetcode
+
+import "container/heap"
+
+func eatenApples(apples []int, days []int) int {
+	h := hp{}
+	i := 0
+	var ans int
+	for ; i < len(apples); i++ {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if apples[i] > 0 {
+			heap.Push(&h, data{apples[i], i + days[i]})
+		}
+		if len(h) > 0 {
+			minData := heap.Pop(&h).(data)
+			ans++
+			if minData.left > 1 {
+				heap.Push(&h, data{minData.left - 1, minData.end})
+			}
+		}
+	}
+	for len(h) > 0 {
+		for len(h) > 0 && h[0].end <= i {
+			heap.Pop(&h)
+		}
+		if len(h) == 0 {
+			break
+		}
+		minData := heap.Pop(&h).(data)
+		nums := min(minData.left, minData.end-i)
+		ans += nums
+		i += nums
+	}
+	return ans
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+type data struct {
+	left int
+	end  int
+}
+
+type hp []data
+
+func (h hp) Len() int           { return len(h) }
+func (h hp) Less(i, j int) bool { return h[i].end < h[j].end }
+func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
+
+func (h *hp) Push(x interface{}) {
+	*h = append(*h, x.(data))
+}
+
+func (h *hp) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+```